Changing a random number into a Key Code

Question

lOkay so I have gotten stumped on this one...

What i'm trying to achieve is to generate a series of numbers between 0-9. After The generation of the numbers, i want them to be typed into a text field on the screen. I have gotten all the essentials working except the changing a random number into a key code. Currently when i go to convert it to a key code, it turns into 0.

Here is my code:

public void random() {
    int rand = new Random().nextInt(10);
            System.out.println(""+rand);
    convert();
}

public void convert() {
    switch (rand) {
    case 0:
        rand = KeyEvent.VK_0;
                    System.out.println(""+0);
        break;
    case 1:
        rand = KeyEvent.VK_1;
                    System.out.println(""+1);
        break;
    case 2:
        rand = KeyEvent.VK_2;
                    System.out.println(""+2);
        break;
    case 3:
        rand = KeyEvent.VK_3;
                    System.out.println(""+3);
        break;
    case 4:
        rand = KeyEvent.VK_4;
                    System.out.println(""+4);
        break;
    case 5:
        rand = KeyEvent.VK_5;
                    System.out.println(""+5);
        break;
    case 6:
        rand = KeyEvent.VK_6;
                    System.out.println(""+6);
        break;
    case 7:
        rand = KeyEvent.VK_7;
                    System.out.println(""+7);
        break;
    case 8:
        rand = KeyEvent.VK_8;
                    System.out.println(""+8);
        break;
    case 9:
        rand = KeyEvent.VK_9;
                    System.out.println(""+9);
        break;
    }
}

The out put is this:

5
0

What it should be:

5
53

Any suggestions/solutions?

Thanks!!

`System.out.printLn`? I though it was `System.out.println`, are you sure this code compiles and run? — Luiggi Mendoza, Oct 23 '12 at 06:07
And why are you prepending "" before each value? Just `System.out.println(5)` will work fine... It's not clear how you expect two digits to be printed out for the second line at all, btw... — Jon Skeet, Oct 23 '12 at 06:08
The rand variable is not defined in the convert method. If it is a class member then the random method does not assign the value to the right variable. — mkhelif, Oct 23 '12 at 06:09
Also, I wonder how your `rand` variable goes from the `random` method to the `convert` method w/o being passed as parameter and not being an attribute of your class. — Luiggi Mendoza, Oct 23 '12 at 06:10
Also, you declared your `rand` variable in your random method which is not passed to convert method. I think that should not even compile, unless you have declared an instance variable. — Rohit Jain, Oct 23 '12 at 06:10
Okay, for all the people asking, i have the rand declared before the random() method as: int rand; — user1767369, Oct 23 '12 at 06:14
Your last comment shows that you don't even know how your code behaves when you run it. I recommend you to use a debugger and see what your actual code does. When you have a real question, come back and we gladly help you. — Luiggi Mendoza, Oct 23 '12 at 06:15
I know what my code is SUPPOSED to do, but the reason i am asking for help is because it is not doing that the way i have set it up. All I'm asking for is some guidance in the right direction to know why my switch statement is changing all the numbers into 0 instead of the respected key code. If you want me to post all of my code i will gladly do that for you, that is if it will better aid you in understanding what i need help with. — user1767369, Oct 23 '12 at 06:19
Are you sure you're not [shadowing](http://stackoverflow.com/q/1092099/1065197) a `rand` attribute in your class with the `int rand` local variable inside the `random` method (because that's how it looks and that's what other people is saying to you in comments/answers). — Luiggi Mendoza, Oct 23 '12 at 06:23

Azodious · Answer 1 · 2012-10-23T06:31:32.340

2

I think you have a global rand variable intialized to 0.

Do following changes and it should work fine:

Change signature of convert to following:
```
public void convert(int rand) 
```
While calling convert from random,call as
```
convert(rand);
```

EDIT:

Based on your comment, that you've int rand declared above random() method. You need not declare rand again.

Change following line

int rand = new Random().nextInt(10); // a new rand variable is created local to method

to

rand = new Random().nextInt(10); // already declared rand variable is used.

edited Oct 23 '12 at 06:31

answered Oct 23 '12 at 06:11

Azodious

13,752
1
36
71

Convert method doesn't call random. where are you seeing recursion? – Azodious Oct 23 '12 at 06:30
Oh .. ya .. corrected. thanks. but it would be a compiler error if it was a method name. – Azodious Oct 23 '12 at 06:31
Okay i have changed to your edits and it converts them to the right codes. Now my new problem is the error: Exception in thread "AWT-EventQueue-0" java.lang.IllegalArgumentException: Invalid key code at sun.awt.windows.WRobotPeer.keyPress(Native Method) at java.awt.Robot.keyPress(Unknown Source) and 55 (the one that was supposed to be the input) is a correct code... – user1767369 Oct 23 '12 at 06:35

Tobias Ritzau · Answer 2 · 2012-10-23T06:15:49.240

1

What you do seems really strange. Is the text field in your own app? Anyway, the way you get 0 is because you must use an instance variable (rand) in the convert method and you assign the random number to a local variable (also called rand but still a different variable). Pass the random variable as a parameter to the convert method like this:

public void random() {
    int rand = new Random().nextInt(10);
    System.out.println(rand);
    convert(rand);
}

public void convert(int rand) {
    switch (rand) {
    ...

Then change the prints to System.out.println(rand);

edited Oct 23 '12 at 06:15

answered Oct 23 '12 at 06:10

Tobias Ritzau

3,327
2
18
29

The text field is not in my own app, I'm using the Robot class to input it in. – user1767369 Oct 23 '12 at 06:13
You should also store the Random object into an instance variable. To get a good random sequence you should use the same object for all random numbers. – Tobias Ritzau Oct 23 '12 at 06:17

score 0 · Answer 3 · answered Oct 23 '12 at 06:16

Your rand is declared local to your random method

int rand = new Random().nextInt(10);

But you are referencing it from within your convert method

switch (rand) {

This suggests that rand has already being declared as an instance variable (otherwise it won't compile) and you are ignoring it...

score 0 · Accepted Answer · answered Oct 23 '12 at 06:32

Luiggi is right. That code does not compile.

Anyway, if I where to guess, this code might work for you:

public void random() {
    int rand = new java.util.Random().nextInt(10);
    convert(rand);
}

public void convert(int rand) {
    int vk = 0;
    switch (rand) {
    case 0:
        vk = java.awt.event.KeyEvent.VK_0;
        break;
    case 1:
        vk = java.awt.event.KeyEvent.VK_1;
        break;
    case 2:
        vk = java.awt.event.KeyEvent.VK_2;
        break;
    case 3:
        vk = java.awt.event.KeyEvent.VK_3;
        break;
    case 4:
        vk = java.awt.event.KeyEvent.VK_4;
        break;
    case 5:
        vk = java.awt.event.KeyEvent.VK_5;
        break;
    case 6:
        vk = java.awt.event.KeyEvent.VK_6;
        break;
    case 7:
        vk = java.awt.event.KeyEvent.VK_7;
        break;
    case 8:
        vk = java.awt.event.KeyEvent.VK_8;
        break;
    case 9:
        vk = java.awt.event.KeyEvent.VK_9;
        break;
    }
    System.out.println(rand+"->"+vk);
}

Note that since VK_0 through VK_9 has values 48 through 57 so you might as well do this:

public void randomAndConvert() {
    int rand = new java.util.Random().nextInt(10);
    int vk = rand+48;
    System.out.println(rand+"->"+vk);
}

Finally, a couple of things about the original code:

printLn should be println
rand is never declared in convert
rand is never passed to convert throught parameter
rand is never printed out in convert

Hope this will lead you down the right path :)

Thanks for this! I don't know why you would get compile errors (I haven't) but will change to this just in-case... The printLn(); was just a typo, was copying off my laptop to post on my desktop (sorrys!)... — user1767369, Oct 23 '12 at 06:41
ok. I guess you have `rand` declared somewhere as mentioned by others. Then the code should "work". I'm not a big fan of using class members for method communication, but ymmv :) — Thobias Bergqvist, Oct 23 '12 at 07:06

Changing a random number into a Key Code

4 Answers4